Euler totient and Mobius function

[Alhazen]

Hi,

Here is an identity to prove :

rightformula.GIF (3.55 KiB) Viewed 7996 times

Is there a way to express mu(n) as function of phi(n)?

Thank you for any clue

Ps : I'm sorry I corrected the mistake.

[Alhazen]

Is it a new result or is it yet known?
Thank you for any clue.

[thundre]

1: 1 0
2: 3 1
3: 6 1

I don't see them ever being equal. Why even sum the right side to n? floor(n/(n+1)) = 0 for all positive n, so any expression multiplied by that will be 0.

[Alhazen]

1: 1 0
2: 3 1
3: 6 1

I don't see them ever being equal. Why even sum the right side to n? floor(n/(n+1)) = 0 for all positive n, so any expression multiplied by that will be 0.

You are right I did a big mistake.
It is k not k+1

Sorry sorry sorry.

[thundre]

It is k not k+1

The updated formula checks out.

a(n) seems to approach n² * .60792 for large n.

I have no idea how to prove the right and left sides are always equal.

[Alhazen]

Thank you very much and once again sorry my mistake.

[kevinsogo]

ceil(n/k) is just floor(n/k) + 1 unless k divides n, so the right hand side is equal to:

$$\sum_{k=1}^n \mu\left(k\right) \left\lfloor \frac{n}{k} \right\rfloor \left(\left\lfloor \frac{n}{k} \right\rfloor + 1\right) - \sum_{k|n} \mu\left(k\right) \left\lfloor \frac{n}{k} \right\rfloor$$
$$= \sum_{k=1}^n \mu\left(k\right) \left\lfloor \frac{n}{k} \right\rfloor^2 + \sum_{k=1}^n \mu\left(k\right) \left\lfloor \frac{n}{k} \right\rfloor - \sum_{k|n} \mu\left(k\right) \left(\frac{n}{k}\right)$$
$$= S_1 + S_2 - S_3 $$

Now, we can prove the following:
S₁ = 2 [sum]_1[le]k[le]n φ(k) - 1
S₂ = 1
S₃ = φ(n)
Thus it is equal to your left hand side.

The proofs of these are relatively simple so you may try proving them yourself.

Hints:
The proof for S₁ and S₂ uses concepts from Problem 351 (View Problem).
The proof for S₃ uses the substitution l = n/k and Möbius inversion.

[Alhazen]

Thank you for the proof.
No one found it in another forum.
Someone had found part of it.

[Alhazen]

It is k not k+1

The updated formula checks out.

a(n) seems to approach n² * .60792 for large n.

I have no idea how to prove the right and left sides are always equal.

And 0.60792 = 6/(π^2)

[Alhazen]

6/(π^2) is the density of squarefree numbers.
What can we do with that?

[Lord_Farin]

6/(π^2) is the density of squarefree numbers.
What can we do with that?

The squarefree numbers are the ones for which the Möbius function is nonzero.

[Alhazen]

6/(π^2) is the density of squarefree numbers.
What can we do with that?

The squarefree numbers are the ones for which the Möbius function is nonzero.

Yes! You are right.
Did someone said otherwise?
There are some links to exploit.
Did you take a look to my new formulation of the Mertens function ?

[Ethan1024]

From the Dirichlet convolution representation of the totient function

k3.PNG (1.14 KiB) Viewed 6645 times

Though it still sort of feels like I am cheating since I am still using lower order values of the mobius function,
and because it requires knowing all the divisors of n.